Problem: Factor the following expression: $x^2 + 9x + 18$
Explanation: When we factor a polynomial, we are basically reversing this process of multiplying linear expressions together: $ \begin{eqnarray} (x + a)(x + b) &=& xx &+& xb + ax &+& ab \\ \\ &=& x^2 &+& {(a + b)}x &+& {ab} \end{eqnarray} $ $ \begin{eqnarray} \hphantom{(x + a)(x + b) }&\hphantom{=}&\hphantom{ xx }&\hphantom{+}&\hphantom{ (a + b)x }&\hphantom{+}& \\ &=& x^2 & +& {9}x& +& {18} \end{eqnarray} $ The coefficient on the $x$ term is $9$ and the constant term is $18$ , so to reverse the steps above, we need to find two numbers that add up to $9$ and multiply to $18$ You can try out different factors of $18$ to see if you can find two that satisfy both conditions. If you're stuck and can't think of any, you can also rewrite the conditions as a system of equations and try solving for $a$ and $b$ $ {a} + {b} = {9}$ $ {a} \times {b} = {18}$ The two numbers $6$ and $3$ satisfy both conditions: $ {6} + {3} = {9} $ $ {6} \times {3} = {18} $ So we can factor the expression as: $(x + {6})(x + {3})$